Calculate the Length of a Curve: y=(e^x+e^-x)/2 from x=0 to x=ln4

[John O' Meara]

Find the length of the curve y=(e^x+e^-x)/2 from x=0 to x=ln4.
$$\frac{dy}{dx}= \frac{1}{2} \frac{d}{dx}e^x + \frac{1}{2} \frac{d}{dx}e^{-x}= \frac{1}{2}(e^x-e^{-x}) \\$$.
Therefore $$(\frac{dy}{dx})^2=\frac{1}{4}(e^x-e^{-x})\\$$ Therefore $$1+(\frac{dy}{dx})^2=\frac{1}{4}(e^{2x} -2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}\\$$.
Therefore $$\int_0^{ln4}\frac{1}{2}(e^{2x}-2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}dx$$. Do I solve this by substitution by letting $$u^2 = e^{2x}-2e^xe^{-x}+e{-2x}+4$$? Or what way should I approach it? Using the said substitution I get the following integral: $$\int u^2 \frac{1}{e^{2x}-e^{-2x}}du$$

 

[cristo]

It will be a far simpler problem if you use the fact that $$\frac{e^x+e^{-x}}{2}=\cosh x$$. Have you encountered hyperbolic functions?

 

[John O' Meara]

No I haven't encountered hyperbolic functions yet. They are two more chaphers away yet.Thanks for the help.

 

[cristo]

No I haven't encountered hyperbolic functions yet. They are two more chaphers away yet.Thanks for the help.

Well, all you need to know for this question is that $$\frac{d}{dx}\cosh x=\sinh x , \hspace{1cm} \frac{d}{dx}\sinh x=\cosh x$$ and the identity $$\cosh^2x-\sinh^2x=1$$

 

[John O' Meara]

You said it would be a far simpler problem to use the fact that $$\frac{e^x+e^{-x}}{2}=coshx$$. Does that imply that it can still be done without the hyperbolic function method. If it does I would like to see the other longer way of solving this integral, please. Thanks.

 

[Dick]

$$\int_0^{ln4}\frac{1}{2}(e^{2x}-2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}dx$$

Start here. e^x*e^(-x)=1, right?

 

[John O' Meara]

Yes, I got as far as $$\int u^2\frac{1}{e^{x}-e^{-x}}du$$ but I cannot express $$e^x-e^{-x}$$ in terms of u.

 

[Dick]

Then back up. It's much simpler than you are making it out to be. I'm trying to get you to notice that the expression inside of the (1/2) power in my last posting is a PERFECT SQUARE.

 

[John O' Meara]

Yes, But what property of a perfect square am I to notice?

 

[Dick]

That sqrt(a^2)=a. Simplify the integral before you start substituting.

 

[John O' Meara]

$$\frac{1}{4}\int_0^{In4}((e^x + e^{-x}})^2)^\frac{1}{2}dx$$, is the integral to be solved.

 

[Dick]

Yes. So you can eliminate the exponents and, voila, you don't need a substitution.

 

[Dick]

PS. You have an extra factor of (1/2) in there.

 

[John O' Meara]

Thanks very much Dick.